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Given two points $p_1,p_2$ on a circular billard table. I want to know all billard trajectories from $p_1$ to $p_2$ hitting the boundary precisely once.

Model of the circular billard:

Denote the orgin of $\mathbb{R}^2$ by $M:=(0,0)$ and assume the billard table has radius one. Thus the boundary of the table is given by $\lbrace (x,y) \in \mathbb{R}^2 : x^2 +y^2 =1 \rbrace$. One can assume that the point $p_1$ is given by $p_1=(0,a)$ for some $a \in (-1,0)$. Notice that the problem is trivial if $p_1$ is either the midpoint $M$ or is lying on the boundary (i.e. $a=0$ or $a=-1$). The situation above can be achieved by rotation.

The point $p_2$ can after rotation and reflection assumed to be given by $p_2=(b,c)$ for some $(b,c) \in (0,1) \times (-1,1)$ satisfying $b^2 + c^2 \leq 1$.

Now the problem is to find for given $a,b,c$ a point $(x,y)$ such that the billard tratjectory from $p_1=(0,a)$ to $(x,y)$ is going through $p_2=(b,c)$. (Billard trajectory means that the angle of incidence is equal to the angle of reflection).

First approach:

Denote by $B$ the point at the boundary of the billard table such that the trajectory from $p_1$ to $p_2$ goes through $B$. For given $p_1=(0,a), p_2=(b,c)$ the distance from $p_1$ to $B$ and the distance from $B$ to $p_2$ should be minimal. Therefore to find $B$ define the function

$$f: S_1((0,0)) \rightarrow \mathbb{R}; x \mapsto \Vert p_1 - x \Vert + \Vert p_2 - x \Vert$$

Thus the problem above reduces to finding the minimum of $f$. This problem can be attacked using Lagrange multipliers.

Thus one can assume $f$ is defined an all of $\mathbb{R}^2$. Its gradient is then given by

$$ \nabla f(x,y) = \begin{pmatrix} \frac{x}{\sqrt{x^2 + (y-a)^2}} + \frac{x-b}{\sqrt{(x-b)^ 2+ (y-c)^2}}\\ \frac{y-a}{\sqrt{x^2 + (y-a)^2}} + \frac{y-c}{\sqrt{(x-b)^2 + (y-c)^2}} \end{pmatrix}. $$

At this point I get stuck, since calculations seem to get to difficult.

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  • $\begingroup$ Are your billiard balls (infinitely small) points instead of finite-sized spheres, then? $\endgroup$ – amd Feb 18 '18 at 0:25
  • $\begingroup$ Yes the billards balls are rather particles than spheres with definite size. $\endgroup$ – Near Feb 18 '18 at 9:27
  • $\begingroup$ There can be up to four solutions, hence you must solve a 4th degree equation, in general. The solutions of such an equation can be very large expressions if you want them expressed as functions of parameters $a$, $b$, $c$. $\endgroup$ – Aretino Feb 18 '18 at 12:07
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    $\begingroup$ This might be of interest: unige.ch/~gander/Preprints/Billiard.pdf $\endgroup$ – Steve Kass Feb 18 '18 at 21:22
  • $\begingroup$ @SteveKass Thank you for pointing out the reference. I am currently having a look at it and is very interesting. $\endgroup$ – Near Mar 4 '18 at 10:39
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I prefer to choose coordinate axes such that points $p_1$ and $p_2$ have the same $y$: $$ p_1=(a,d),\quad p_2=(b,d). $$ Let $M=(0,0)$ and $V=(x,y)$ be the bouncing point ($x^2+y^2=1$), so that $VM$ is the bisector of angle $\angle p_1Vp_2$. If $N$ is the intersection between $VM$ and $p_1p_2$, then: $N=(dx/y,d)$. On the other hand, from the bisector theorem we have $Vp_1:Vp_2=Np_1:Np_2$, that is: $$ [(x-a)^2+(y-d)^2](by-dx)^2=[(x-b)^2+(y-d)^2](ay-dx)^2. $$ This equation has the spurious solution $y=d$: factoring that out and performing some simplification leads to: $$ 2d(a+b)y^2-2(d^2-ab)xy+2dx-(a+b)y-(a+b)d=0. $$ Together with $x^2+y^2=1$, this equation forms a fourth order system, which can have up to four real solutions. On the other hand, one can easily find some examples where four solutions actually exist, for instance: $a = -0.7$, $b = 0.6$, $d = 0.1$.

An explicit solution in terms of $a$, $b$, $d$ is too large to be of any use. You'd better substitute into the equation the values of the parameters before solving it.

EDIT.

The above equation represents a hyperbola and the bouncing points are then the intersections between this hyperbola and the unit circle. They can thus be found with GeoGebra: https://www.geogebra.org/m/WW3KkeJB.

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  • $\begingroup$ Richter-Gebert’s Perspectives on Projective Geometry has an algorithm for computing the intersections of two conics. It does involve solving a cubic equation, but the text includes a straightforward, though tedious, algorithm for solving that particular equation. This algorithm is discussed in this answer. $\endgroup$ – amd Feb 25 '18 at 8:55
  • $\begingroup$ @amd Thank you for pointing me to that algorithm. However I don't see a clear advantage in making all that work only to switch from a quartic equation to a cubic. $\endgroup$ – Aretino Feb 25 '18 at 9:52
  • $\begingroup$ Thank you for the geogebra applet. It is very nice. As you pointed out to compute the general solutions is tedious. Can one maybe say more about a specific trjacetory. Lets say the one with minimal length? $\endgroup$ – Near Mar 4 '18 at 10:38

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